How do you evaluate the limit #1/x^2# as x approaches #0#?

Answer 1

#lim_(x->0) 1/x^2 = +oo#

This is quite evident, since, for #x->0#, #x^2# is positive and indefinitely small, so its reciprocal is positive and indefinitely large.
However, it can be proved easily in the #delta-epsilon# form:
GIven any #M > 0# we can choose #delta_M = 1/sqrt(M)#.
So, for #x in (-delta_M, delta_M)# we have:
#abs(x) < 1/sqrt(M)#
#x^2 < 1/M#
#1/x^2 > M#

which proves the point.

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Answer 2

To evaluate the limit of 1/x^2 as x approaches 0, we substitute 0 into the expression for x. This gives us 1/0^2, which simplifies to 1/0. However, division by zero is undefined in mathematics. Therefore, the limit of 1/x^2 as x approaches 0 does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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